Computing: $ \lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2} $ $$
\lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}
$$
The answer is 0. Cannot seem to understand how the answer is 0.I know that the first part is 0 but I'm confused on how to deal with the natural log?
Why is squeeze theorem not a good approach? $0<|(x^5)<(x^2)^2|=|x|$ so $|x^5+y^5|<|(x^2+y^2)^2|$ so therefor 
$$0< \dfrac{|x^5+y^5|}{|(x^2+y^2)^2|}<1$$ then multiply both sides with $\ln(x^2+y^2)$. and take the $$\lim_{(x,y)\to (0,0)} \ln(x^2+y^2).$$ and since it is not $0$ but its negative infinity the limit doesn't exist by the squeeze theorem. which is not the right answer.
 A: Hint. Note that by letting $x=\rho\cos(\theta)$ and $y=\rho\sin(\theta)$, we have that as $(x,y)\to(0,0)$ then $\rho\to 0$ and
$$0\leq \left|\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}\right|\leq
\frac{(|\rho\cos\theta|^5+|\rho\sin\theta|^5)|\ln(\rho^2)|}{\rho^4}
\leq \frac{\rho^5(1+1)|\ln(\rho^2)|}{\rho^4}={4\rho|\ln(\rho)|}.$$
Can you take it from here? 
A: For all norm we have $|x|, |y|\le \|(x,y)\|$
then we get $$|x^5+ y^5|\le 2\|(x,y)\|^5$$
Since $\lim_{x\to 0}x\ln x =  0$ and $\ln(x^2+y^2) = 2\ln\|(x,y)\| $  we obtain,
$$\left|\lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}\right| \le \lim_{\|(x,y)\|\to 0} 4\|(x,y)\|\left|\ln\|(x,y)\|\right| =0$$
A: Okay, let's do this properly: We use polar coordinates $x=r\cos \phi$ and $y=r\sin \phi$. We will try to use the sandwich (or squeeze) theorem:
\begin{align}
0&\leq \left|\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}\right| \\
 &= \frac{r^5|\cos\phi + \sin\phi|\cdot|\ln r^2|}{r^4} \\
 &\leq r\cdot(1+1)\cdot2\cdot|\ln r| = 4r |\ln r|.
\end{align}
We now need to show that $r|\ln r|\to 0$. For this we will use l'hospital's theorem (note that $|\ln r|=-\ln r$ for small $r$):
\begin{align}
&\lim_{r\to 0} r |\ln r| = \lim_{r\to 0} \frac{-\ln r}{\frac1r} = “-\frac\infty\infty”\\
&\lim_{r\to 0} \frac{-\frac1r}{-\frac1{r^2}} = \lim_{r\to0} r = 0.
\end{align}
Since the latter limit converges, both limits are equal and therefore $r|\ln r|\to 0$ for $r\to 0$. And thusly, by the sandwich theorem, the limit you asked for converges to $0$.
